forcesrotational-dynamics
Before rotation is introduced into the course, Atwood machines were considered to have equal tension throughout. In this particular problem however, the pulley did have a significant body which rotated about an axis. In captions the authors stated that the tensions may not be equal even though it is on the same rope, an indeed the tensions are different numerically.
Why can we assume equal tensions when the pulley is negligible but not when it is significant?
It's alright, tension is pretty subtle! Let me answer your questions a bit out of order.
- If you pick any and all points on the rope, would there be two opposing tensions at every one of those points?
- Is tension uniform throughout the rope?
You can think about the rope as a lot of tiny masses connected together by springs; this is a cheap approximation for how tension works on the atomic level, where the springs are stretching chemical bonds and the masses are atoms.
In our simple model of tension, every atom is pulled by a spring on the left and a spring on the right, with forces $T_1$ and $T_2$. Then by Newton's second law, $$T_1 - T_2 = ma$$ where $m$ is the mass of the atom. Since $m$ is very very tiny compared to the other masses in the problem, we must have $$T_1 \approx T_2.$$ Applying this to every mass, we conclude that each little spring / chemical bond has approximately the same tension, so we can simply talk about "the tension in the rope". This is a good approximation as long as the total mass of the rope is much smaller than the masses of the blocks.
How might differences in mass between object A and object B (which, sorry if the diagram was misleading in the sizes, can have any mass) play into the tension?
Well, now that we've established that there's a uniform tension $T$, we need to figure out what that tension is. The constraint here is that the rope is taut, which means that it can't be scrunching up or stretching out; that translates to the constraint that the accelerations of the two blocks are equal in magnitude. This equation determines the tension.
How does the relationship between the force of gravity on mass B and the tension in the rope play into this? Isn't the tension caused by that force of gravity? Doesn't that mean that if tensions cancel, the force of gravity's effect is canceled as well? (I have also been told that the blocks will have the same magnitude of acceleration. Why is this?)
No, the tension isn't equal to the weight of block B, it's whatever is necessary to satisfy the above constraint. For example, suppose that block B was very very heavy, so the acceleration of the whole system will be close to $g$. In this case, the tension is actually quite small compared to the weight of block B, because you only need a little tension to make block A have the same acceleration.
(In fact, as block B gets infinitely heavy, you can show that the tension doesn't go to infinity -- instead, it becomes the weight of block A! It's neat to try to prove this, and see how it works.)
Does the pulley affect tensions? For example, we know that there's a positive tension affecting mass A. Is there still a positive tension in existence on the other side of the pulley, or just the negative tension that's acting on B? Might there be some sort of effect whereby two sets of opposite tensions, one set on each side of the pulley, cancel each other out?
This is a little tricky to word. The tensions of the two tiny springs attached to each atom approximately cancel out, as shown above. But that doesn't mean that the tension is zero -- all of those springs are still stretched.
Since the pulley is frictionless, it doesn't have any effect except that it 'turns around' the tension. You can show this by considering the three forces on each atom (two springs, one normal force) which gives $T_1 \approx T_2$ as before.
Anyway, I've been told that the net force is the force of gravity on mass B (the hanging one). But I don't really understand why, even after extensive discussion with various people.
That's not your fault, the question is just worded badly. There are lots of forces involved in this problem acting on different things: gravity on both blocks, normal on one block, and normal from the pulley on the rope. It's not very clear what "the" net force even means.
I will only consider taut, inextensible ropes. This represents an approximation regime where tension is much less than the Young's modulus of the rope material times the cross-sectional area of the rope. (As we will see this condition is enough to guarantee the conclusion for straight segments of the rope as long as they are light compared to what ever they are attached to.)
Let's start with the easiest case: there is a frame of reference in which the entire rope is still.
In this case we can reason thus: the rope is subject to net zero external force (by Newton's 2nd Law). Further any internal element of the rope must be subject to equal and opposite forces from neighboring elements. Those forces are the tension in the rope so the tension is the same throughout.
[Note that the rope need not be massless.]
Next easiest case is the rope is in motion with constant speed along it's own length, but it may pass over circular pulleys with no friction at bearing (but with enough friction at the grove that the rope does not slip) and the like so any given part of the rope might change directions at times.
The argument in part (1) applies to segments between the pulleys.
The pulleys themselves have no angular acceleration and are therefore subject to zero net torque. So, the segments on either side of the pulley have the same tension because they have to exert equal and opposite torques.
[Note that the neither the rope nor the pulleys need to be massless.]
Now we come to cases where the rope is accelerating along it's own length.
If we assume a massless rope then any change of tension along the length would cause arbitrary acceleration. That's a nasty case because it s non-physical, but it represents an approximation of a regime where the mass of the rope is much smaller than the mass of anything it is connected to. Demonstration Atwood's machines are roughly like that.
A straight, massive rope (segment) accelerating along it's own length. The segment is subject to a net external force $F_n = F_1 - F_2$ where $F_1$ and $F_2$ are the net forces at either end, so that the acceleration is $a = F_n/m$. The rope segment is assumed to have uniform mass density $\lambda = m/l$. At any fraction $f$ along the length of the rope the tension must be such that the segment on each side of conceptual divide has the same acceleration (to prevent extension without allowing slackness). So tension $\tau$ (which is the same in each direction from Newton's 3rd Law) at position $fl$ must meet the requirements $$ a_1 = \frac{F_1 - \tau}{\lambda fl} \;,$$ and $$ a_2 = \frac{\tau - F_2}{\lambda (1-f)l} \;,$$ where $a_i$ is the acceleration of the segment closer to force $F_i$, and $a_1 = a_2$ is required. Thus \begin{align*} \frac{F_1 - \tau}{\lambda fl} &= \frac{\tau - F_2}{\lambda (1-f)l} \\ \frac{F_1 - \tau}{f} &= \frac{\tau - F_2}{1-f} \\ (1 - f)(F_1 - \tau) &= f(\tau - F_2) \\ -\tau &= (f - 1)F_1 - fF_2 \\ \tau &= F_1 - fF_n \;. \end{align*} Now, this depends on $f$, which means that it is not constant, right?
OK, but the term is $fF_n$, and $F_n = ma$ where $m$ is the mass of the rope segment. So, if the rope is much less massive than the things attached to it's ends, this terms could be negligible compared to $F1$ (or, indeed $F_2$ since the labeling is arbitrary). Then we get $$ \tau \approx F_1 \,,$$ which justifies the claim in item (3) that the hypothetical "massless" condition is an approximation for "very light".
Next we could let these things run over low friction pulleys. We can recover the 'tension the same throughout' conclusion only if the pulleys are also "very light" so that we can use an argument like that in item (2) because the torque on them will arise from force small compared to $F_1$ or $F_2$.
Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light.
Best Answer
It is all a matter of progression.
When one starts studying Mechanics it is in terms of point masses, massless and inextensible strings, frictionless and massless pulleys, etc..
This is done to produce some foundations on which to build.
Strings and pulleys are just devices for transferring forces from one place to another and changing the directions of forces.
A step forward is to introduce rotational dynamics where bodies are no longer treated as point masses and pulleys suddenly become nearer to those that can be found in the real world.
It appears that you have taken such a step.
If you have a pulley of moment of inertia $I_{\rm C}$ about its centre of mass then to produce an angular acceleration $\alpha$ of the pulley requires a torque about the centre of mass of the pulley of $\tau = I_{\rm C} \;\alpha$.
Note that if the pulley is massless then it has no moment of inertia and so requires no torque to accelerate it.
The torque is applied using a string which can still be assumed massless and inextensible but now communicates with the pulley via a static frictional force.
This now means that the tension in the string is no longer constant as is shown in the solution to the capstan problem. (However note that there are no accelerations in the derivation).
You normally do not worry about the detail of the interaction between the pulley and the string what interests you is the fact that the tensions at either end of the string are difference and it is that difference which enable the string to apply a torque on the pulley.
In some cases you could assume that the moment of inertia of the pulley is so small as to make the very little difference to the tension in the string which is mirroring the assumption that the pulley is massless.
I wonder what assumptions (simplifications) NASA made when designing their pulley system?